面试必看的sql练习题大全

约定不等于承诺〃 2021-12-22 11:55 297阅读 0赞

# 习题来源于网络，sql语句是自己写的，部分有参考。欢迎指正。本文将持续进行更新 #

# 建立数据库和插入基础数据的sql语句： #

-- ----------------------------
    -- Table structure for course
    -- ----------------------------
    DROP TABLE IF EXISTS `course`;
    CREATE TABLE `course`  (
      `c_id` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL,
      `c_name` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL DEFAULT '',
      `t_id` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL,
      PRIMARY KEY (`c_id`) USING BTREE
    ) ENGINE = InnoDB CHARACTER SET = utf8mb4 COLLATE = utf8mb4_general_ci ROW_FORMAT = Dynamic;
    
    -- ----------------------------
    -- Records of course
    -- ----------------------------
    INSERT INTO `course` VALUES ('01', '语文', '02');
    INSERT INTO `course` VALUES ('02', '数学', '01');
    INSERT INTO `course` VALUES ('03', '英语', '03');
    
    -- ----------------------------
    -- Table structure for score
    -- ----------------------------
    DROP TABLE IF EXISTS `score`;
    CREATE TABLE `score`  (
      `s_id` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL,
      `c_id` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL,
      `s_score` int(3) NULL DEFAULT NULL,
      PRIMARY KEY (`s_id`, `c_id`) USING BTREE
    ) ENGINE = InnoDB CHARACTER SET = utf8mb4 COLLATE = utf8mb4_general_ci ROW_FORMAT = Dynamic;
    
    -- ----------------------------
    -- Records of score
    -- ----------------------------
    INSERT INTO `score` VALUES ('01', '01', 80);
    INSERT INTO `score` VALUES ('01', '02', 90);
    INSERT INTO `score` VALUES ('01', '03', 99);
    INSERT INTO `score` VALUES ('02', '01', 70);
    INSERT INTO `score` VALUES ('02', '02', 60);
    INSERT INTO `score` VALUES ('02', '03', 80);
    INSERT INTO `score` VALUES ('03', '01', 80);
    INSERT INTO `score` VALUES ('03', '02', 80);
    INSERT INTO `score` VALUES ('03', '03', 80);
    INSERT INTO `score` VALUES ('04', '01', 50);
    INSERT INTO `score` VALUES ('04', '02', 30);
    INSERT INTO `score` VALUES ('04', '03', 20);
    INSERT INTO `score` VALUES ('05', '01', 76);
    INSERT INTO `score` VALUES ('05', '02', 87);
    INSERT INTO `score` VALUES ('06', '01', 31);
    INSERT INTO `score` VALUES ('06', '03', 34);
    INSERT INTO `score` VALUES ('07', '02', 89);
    INSERT INTO `score` VALUES ('07', '03', 98);
    
    -- ----------------------------
    -- Table structure for student
    -- ----------------------------
    DROP TABLE IF EXISTS `student`;
    CREATE TABLE `student`  (
      `s_id` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL,
      `s_name` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL DEFAULT '',
      `s_birth` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL DEFAULT '',
      `s_sex` varchar(10) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL DEFAULT '',
      PRIMARY KEY (`s_id`) USING BTREE
    ) ENGINE = InnoDB CHARACTER SET = utf8mb4 COLLATE = utf8mb4_general_ci ROW_FORMAT = Dynamic;
    
    -- ----------------------------
    -- Records of student
    -- ----------------------------
    INSERT INTO `student` VALUES ('01', '赵雷', '1990-01-01', '男');
    INSERT INTO `student` VALUES ('02', '钱电', '1990-12-21', '男');
    INSERT INTO `student` VALUES ('03', '孙风', '1990-05-20', '男');
    INSERT INTO `student` VALUES ('04', '李云', '1990-08-06', '男');
    INSERT INTO `student` VALUES ('05', '周梅', '1991-12-01', '女');
    INSERT INTO `student` VALUES ('06', '吴兰', '1992-03-01', '女');
    INSERT INTO `student` VALUES ('07', '郑竹', '1989-07-01', '女');
    INSERT INTO `student` VALUES ('08', '王菊', '1990-01-20', '女');
    
    -- ----------------------------
    -- Table structure for teacher
    -- ----------------------------
    DROP TABLE IF EXISTS `teacher`;
    CREATE TABLE `teacher`  (
      `t_id` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL,
      `t_name` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL DEFAULT '',
      PRIMARY KEY (`t_id`) USING BTREE
    ) ENGINE = InnoDB CHARACTER SET = utf8mb4 COLLATE = utf8mb4_general_ci ROW_FORMAT = Dynamic;
    
    -- ----------------------------
    -- Records of teacher
    -- ----------------------------
    INSERT INTO `teacher` VALUES ('01', '张三');
    INSERT INTO `teacher` VALUES ('02', '李四');
    INSERT INTO `teacher` VALUES ('03', '王五');

# 以下为sql问题以及答案： #

-- 1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数 
    	
    SELECT
    	a.*,
    	b.s_score AS 01 _score,
    	c.s_score AS 02 _score 
    FROM
    	student a
    	JOIN score b ON a.s_id = b.s_id 
    	AND b.c_id = '01'
    	LEFT JOIN score c ON a.s_id = c.s_id 
    	AND c.c_id = '02' 
    	OR c.c_id = NULL 
    WHERE
    	b.s_score > c.s_score
    	
    -- 也可以这样写
    SELECT
    	a.*,
    	b.s_score AS 01 _score,
    	c.s_score AS 02 _score 
    FROM
    	student a,
    	score b,
    	score c 
    WHERE
    	a.s_id = b.s_id 
    	AND a.s_id = c.s_id 
    	AND b.c_id = '01' 
    	AND c.c_id = '02' 
    	AND b.s_score > c.s_score
    			
    -- 2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数
    	
    SELECT
    	a.*,
    	b.s_score AS 01 _score,
    	c.s_score AS 02 _score 
    FROM
    	student a
    	LEFT JOIN score b ON a.s_id = b.s_id 
    	AND b.c_id = '01' 
    	OR b.c_id =
    	NULL JOIN score c ON a.s_id = c.s_id 
    	AND c.c_id = '02' 
    WHERE
    	b.s_score < c.s_score
    			
    
    -- 3、查询平均成绩大于等于60分的同学的学生编号和学生姓名和平均成绩
    SELECT
    	b.s_id,
    	b.s_name,
    	ROUND( AVG( a.s_score ), 2 ) AS avg_score 
    FROM
    	student b
    	JOIN score a ON b.s_id = a.s_id 
    GROUP BY
    	b.s_id,
    	b.s_name 
    HAVING
    	avg_score >= 60;
    	
    
    -- 4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩
    		-- (包括有成绩的和无成绩的)
    		
    SELECT
    	b.s_id,
    	b.s_name,
    	ROUND( AVG( a.s_score ), 2 ) AS avg_score 
    FROM
    	student b
    	LEFT JOIN score a ON b.s_id = a.s_id 
    GROUP BY
    	b.s_id,
    	b.s_name 
    HAVING
    	avg_score < 60 UNION
    SELECT
    	a.s_id,
    	a.s_name,
    	0 AS avg_score 
    FROM
    	student a 
    WHERE
    	a.s_id NOT IN ( SELECT DISTINCT s_id FROM score );
    
    
    -- 5、查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩
    SELECT
    	a.s_id,
    	a.s_name,
    	count( b.c_id ) AS sum_course,
    	sum( b.s_score ) AS sum_score 
    FROM
    	student a
    	LEFT JOIN score b ON a.s_id = b.s_id 
    GROUP BY
    	a.s_id,
    	a.s_name;
    			
    			
    -- 6、查询"李"姓老师的数量 
    SELECT
    	count( t_id ) 
    FROM
    	teacher 
    WHERE
    	t_name LIKE '李%';
    	
    -- 7、查询学过"张三"老师授课的同学的信息 
    SELECT
    	a.* 
    FROM
    	student a
    	JOIN score b ON a.s_id = b.s_id 
    WHERE
    	b.c_id IN ( SELECT c_id FROM course WHERE t_id = ( SELECT t_id FROM teacher WHERE t_name = '张三' ) );
    
    -- 8、查询没学过"张三"老师授课的同学的信息 
    SELECT
    	* 
    FROM
    	student c 
    WHERE
    	c.s_id NOT IN (
    SELECT
    	a.s_id 
    FROM
    	student a
    	JOIN score b ON a.s_id = b.s_id 
    WHERE
    	b.c_id IN ( SELECT a.c_id FROM course a JOIN teacher b ON a.t_id = b.t_id WHERE t_name = '张三' ) 
    	);
    -- 9、查询学过编号为"01"并且也学过编号为"02"的课程的同学的信息
    
    SELECT
    	a.* 
    FROM
    	student a,
    	score b,
    	score c 
    WHERE
    	a.s_id = b.s_id 
    	AND a.s_id = c.s_id 
    	AND b.c_id = '01' 
    	AND c.c_id = '02';
    	
    -- 10、查询学过编号为"01"但是没有学过编号为"02"的课程的同学的信息
    			
    SELECT
    	a.* 
    FROM
    	student a 
    WHERE
    	a.s_id IN ( SELECT s_id FROM score WHERE c_id = '01' ) 
    	AND a.s_id NOT IN ( SELECT s_id FROM score WHERE c_id = '02' )
    			
    
    -- 11、查询没有学全所有课程的同学的信息 
    -- @wendiepei的写法
    SELECT
    	s.* 
    FROM
    	student s
    	LEFT JOIN Score s1 ON s1.s_id = s.s_id 
    GROUP BY
    	s.s_id 
    HAVING
    	count( s1.c_id ) < ( SELECT count( * ) FROM course )
    -- @k1051785839的写法
    SELECT
    	* 
    FROM
    	student 
    WHERE
    	s_id NOT IN ( SELECT s_id FROM score t1 GROUP BY s_id HAVING count( * ) = ( SELECT count( DISTINCT c_id ) FROM course ) )
    -- 12、查询至少有一门课与学号为"01"的同学所学相同的同学的信息 
    
    SELECT
    	* 
    FROM
    	student 
    WHERE
    	s_id IN (
    SELECT DISTINCT
    	a.s_id 
    FROM
    	score a 
    WHERE
    	a.c_id IN ( SELECT a.c_id FROM score a WHERE a.s_id = '01' ) 
    	);
    			
    -- 13、查询和"01"号的同学学习的课程完全相同的其他同学的信息 
    --@ouyang_1993的写法
    SELECT
     Student.*
    FROM
     Student
    WHERE
     s_id IN (SELECT s_id FROM Score GROUP BY s_id HAVING COUNT(s_id) = (
        #下面的语句是找到'01'同学学习的课程数
        SELECT COUNT(c_id) FROM Score WHERE s_id = '01'
       )
     )
    AND s_id NOT IN (
     #下面的语句是找到学过‘01’同学没学过的课程，有哪些同学。并排除他们
     SELECT s_id FROM Score
     WHERE c_id IN(
       #下面的语句是找到‘01’同学没学过的课程
       SELECT DISTINCT c_id FROM Score
       WHERE c_id NOT IN (
         #下面的语句是找出‘01’同学学习的课程
         SELECT c_id FROM Score WHERE s_id = '01'
        )
      ) GROUP BY s_id
    ) #下面的条件是排除01同学
    AND s_id NOT IN ('01')
    --@k1051785839的写法
    SELECT
     t3.*
    FROM
     (
      SELECT
       s_id,
       group_concat(c_id ORDER BY c_id) group1
      FROM
       score
      WHERE
       s_id <> '01'
      GROUP BY
       s_id
     ) t1
    INNER JOIN (
     SELECT
      group_concat(c_id ORDER BY c_id) group2
     FROM
      score
     WHERE
      s_id = '01'
     GROUP BY
      s_id
    ) t2 ON t1.group1 = t2.group2
    INNER JOIN student t3 ON t1.s_id = t3.s_id
    
    -- 14、查询没学过"张三"老师讲授的任一门课程的学生姓名 
    
    SELECT
    	a.s_name 
    FROM
    	student a 
    WHERE
    	a.s_id NOT IN (
    SELECT
    	s_id 
    FROM
    	score 
    WHERE
    	c_id = ( SELECT c_id FROM course WHERE t_id = ( SELECT t_id FROM teacher WHERE t_name = '张三' ) ) 
    	);
    
    -- 15、查询两门及其以上不及格课程的同学的学号，姓名及其平均成绩 
    
    SELECT
    	a.s_id,
    	a.s_name,
    	ROUND( AVG( b.s_score ) ) 
    FROM
    	student a
    	LEFT JOIN score b ON a.s_id = b.s_id 
    WHERE
    	a.s_id IN ( SELECT s_id FROM score WHERE s_score < 60 GROUP BY s_id HAVING count( 1 ) >= 2 ) 
    GROUP BY
    	a.s_id,
    	a.s_name
    
    -- 16、检索"01"课程分数小于60，按分数降序排列的学生信息
    
    SELECT
    	a.*,
    	b.c_id,
    	b.s_score 
    FROM
    	student a,
    	score b 
    WHERE
    	a.s_id = b.s_id 
    	AND b.c_id = '01' 
    	AND b.s_score < 60 
    ORDER BY
    	b.s_score DESC;
    		
    -- 17、按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩
    
    SELECT
    	a.s_id,
    	( SELECT s_score FROM score WHERE s_id = a.s_id AND c_id = '01' ) AS 语文,
    	( SELECT s_score FROM score WHERE s_id = a.s_id AND c_id = '02' ) AS 数学,
    	( SELECT s_score FROM score WHERE s_id = a.s_id AND c_id = '03' ) AS 英语,
    	round( avg( s_score ), 2 ) AS 平均分 
    FROM
    	score a 
    GROUP BY
    	a.s_id 
    ORDER BY
    	平均分 DESC;
    			
    -- @喝完这杯还有一箱的写法
    SELECT
    	a.s_id,
    	MAX( CASE a.c_id WHEN '01' THEN a.s_score END ) 语文,
    	MAX( CASE a.c_id WHEN '02' THEN a.s_score END ) 数学,
    	MAX( CASE a.c_id WHEN '03' THEN a.s_score END ) 英语,
    	avg( a.s_score ),
    	b.s_name 
    FROM
    	Score a
    	JOIN Student b ON a.s_id = b.s_id 
    GROUP BY
    	a.s_id 
    ORDER BY
    	5 DESC
    -- 18.查询各科成绩最高分、最低分和平均分：以如下形式显示：课程ID，课程name，最高分，最低分，平均分，及格率，中等率，优良率，优秀率
    -- 及格为>=60，中等为：70-80，优良为：80-90，优秀为：>=90
    SELECT
    	a.c_id,
    	b.c_name,
    	MAX( s_score ),
    	MIN( s_score ),
    	ROUND( AVG( s_score ), 2 ),
    	ROUND(
    	100 * (
    	SUM( CASE WHEN a.s_score >= 60 THEN 1 ELSE 0 END ) / SUM( CASE WHEN a.s_score THEN 1 ELSE 0 END ) 
    	),
    	2 
    	) AS 及格率,
    	ROUND(
    	100 * (
    	SUM( CASE WHEN a.s_score >= 70 AND a.s_score <= 80 THEN 1 ELSE 0 END ) / SUM( CASE WHEN a.s_score THEN 1 ELSE 0 END ) 
    	),
    	2 
    	) AS 中等率,
    	ROUND(
    	100 * (
    	SUM( CASE WHEN a.s_score >= 80 AND a.s_score <= 90 THEN 1 ELSE 0 END ) / SUM( CASE WHEN a.s_score THEN 1 ELSE 0 END ) 
    	),
    	2 
    	) AS 优良率,
    	ROUND(
    	100 * (
    	SUM( CASE WHEN a.s_score >= 90 THEN 1 ELSE 0 END ) / SUM( CASE WHEN a.s_score THEN 1 ELSE 0 END ) 
    	),
    	2 
    	) AS 优秀率 
    FROM
    	score a
    	LEFT JOIN course b ON a.c_id = b.c_id 
    GROUP BY
    	a.c_id,
    	b.c_name
    	
    -- 19、按各科成绩进行排序，并显示排名
    -- mysql没有rank函数
    	select a.s_id,a.c_id,
            @i:=@i +1 as i保留排名,
            @k:=(case when @score=a.s_score then @k else @i end) as rank不保留排名,
            @score:=a.s_score as score
        from (
            select s_id,c_id,s_score from score GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
    )a,(select @k:=0,@i:=0,@score:=0)s
    -- @k1051785839的写法
    (select * from (select 
    t1.c_id,
    t1.s_score,
    (select count(distinct t2.s_score) from score t2 where t2.s_score>=t1.s_score and t2.c_id='01') rank
    FROM score t1 where t1.c_id='01'
    order by t1.s_score desc) t1)
    union
    (select * from (select 
    t1.c_id,
    t1.s_score,
    (select count(distinct t2.s_score) from score t2 where t2.s_score>=t1.s_score and t2.c_id='02') rank
    FROM score t1 where t1.c_id='02'
    order by t1.s_score desc) t2)
    union
    (select * from (select 
    t1.c_id,
    t1.s_score,
    (select count(distinct t2.s_score) from score t2 where t2.s_score>=t1.s_score and t2.c_id='03') rank
    FROM score t1 where t1.c_id='03'
    order by t1.s_score desc) t3)
    
    -- 20、查询学生的总成绩并进行排名
    select a.s_id,
    	@i:=@i+1 as i,
    	@k:=(case when @score=a.sum_score then @k else @i end) as rank,
    	@score:=a.sum_score as score
    from (select s_id,SUM(s_score) as sum_score from score GROUP BY s_id ORDER BY sum_score DESC)a,
    	(select @k:=0,@i:=0,@score:=0)s
    	
    -- 21、查询不同老师所教不同课程平均分从高到低显示 
    		
    SELECT
    	a.t_id,
    	c.t_name,
    	a.c_id,
    	ROUND( avg( s_score ), 2 ) AS avg_score 
    FROM
    	course a
    	LEFT JOIN score b ON a.c_id = b.c_id
    	LEFT JOIN teacher c ON a.t_id = c.t_id 
    GROUP BY
    	a.c_id,
    	a.t_id,
    	c.t_name 
    ORDER BY
    	avg_score DESC;
    -- 22、查询所有课程的成绩第2名到第3名的学生信息及该课程成绩
    			
    			select d.*,c.排名,c.s_score,c.c_id from (
                    select a.s_id,a.s_score,a.c_id,@i:=@i+1 as 排名 from score a,(select @i:=0)s where a.c_id='01'  
    								ORDER BY a.s_score DESC  
                )c
                left join student d on c.s_id=d.s_id
                where 排名 BETWEEN 2 AND 3
                UNION
                select d.*,c.排名,c.s_score,c.c_id from (
                    select a.s_id,a.s_score,a.c_id,@j:=@j+1 as 排名 from score a,(select @j:=0)s where a.c_id='02'  
    								ORDER BY a.s_score DESC
                )c
                left join student d on c.s_id=d.s_id
                where 排名 BETWEEN 2 AND 3
                UNION
                select d.*,c.排名,c.s_score,c.c_id from (
                    select a.s_id,a.s_score,a.c_id,@k:=@k+1 as 排名 from score a,(select @k:=0)s where a.c_id='03' 
    								ORDER BY a.s_score DESC
                )c
                left join student d on c.s_id=d.s_id
                where 排名 BETWEEN 2 AND 3;
    			
    -- 23、统计各科成绩各分数段人数：课程编号,课程名称,[100-85],[85-70],[70-60],[0-60]及所占百分比
    
    
    		select distinct f.c_name,a.c_id,b.`85-100`,b.百分比,c.`70-85`,c.百分比,d.`60-70`,d.百分比,e.`0-60`,e.百分比 from score a
    				left join (select c_id,SUM(case when s_score >85 and s_score <=100 then 1 else 0 end) as `85-100`,
    											ROUND(100*(SUM(case when s_score >85 and s_score <=100 then 1 else 0 end)/count(*)),2) as 百分比
    								from score GROUP BY c_id)b on a.c_id=b.c_id
    				left join (select c_id,SUM(case when s_score >70 and s_score <=85 then 1 else 0 end) as `70-85`,
    											ROUND(100*(SUM(case when s_score >70 and s_score <=85 then 1 else 0 end)/count(*)),2) as 百分比
    								from score GROUP BY c_id)c on a.c_id=c.c_id
    				left join (select c_id,SUM(case when s_score >60 and s_score <=70 then 1 else 0 end) as `60-70`,
    											ROUND(100*(SUM(case when s_score >60 and s_score <=70 then 1 else 0 end)/count(*)),2) as 百分比
    								from score GROUP BY c_id)d on a.c_id=d.c_id
    				left join (select c_id,SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end) as `0-60`,
    											ROUND(100*(SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end)/count(*)),2) as 百分比
    								from score GROUP BY c_id)e on a.c_id=e.c_id
    				left join course f on a.c_id = f.c_id
    				 
    -- 24、查询学生平均成绩及其名次 
    
    		select a.s_id,
    				@i:=@i+1 as '不保留空缺排名',
    				@k:=(case when @avg_score=a.avg_s then @k else @i end) as '保留空缺排名',
    				@avg_score:=avg_s as '平均分'
    		from (select s_id,ROUND(AVG(s_score),2) as avg_s from score GROUP BY s_id ORDER BY avg_s DESC)a,(select @avg_score:=0,@i:=0,@k:=0)b;
    -- 25、查询各科成绩前三名的记录
    			-- 1.选出b表比a表成绩大的所有组
    			-- 2.选出比当前id成绩大的 小于三个的
    SELECT
    	a.s_id,
    	a.c_id,
    	a.s_score 
    FROM
    	score a
    	LEFT JOIN score b ON a.c_id = b.c_id 
    	AND a.s_score < b.s_score 
    GROUP BY
    	a.s_id,
    	a.c_id,
    	a.s_score 
    HAVING
    	COUNT( b.s_id ) < 3 
    ORDER BY
    	a.c_id,
    	a.s_score DESC
    
    -- 26、查询每门课程被选修的学生数 
    
    SELECT
    	c_id,
    	count( s_id ) 
    FROM
    	score a 
    GROUP BY
    	c_id
    
    -- 27、查询出只有两门课程的全部学生的学号和姓名 
    SELECT
    	s_id,
    	s_name 
    FROM
    	student 
    WHERE
    	s_id IN ( SELECT s_id FROM score GROUP BY s_id HAVING COUNT( c_id ) = 2 );
    
    -- 28、查询男生、女生人数 
    SELECT
    	s_sex,
    	COUNT( s_sex ) AS 人数 
    FROM
    	student 
    GROUP BY
    	s_sex
    
    -- 29、查询名字中含有"风"字的学生信息
    
    SELECT
    	* 
    FROM
    	student 
    WHERE
    	s_name LIKE '%风%';
    
    -- 30、查询同名同性学生名单，并统计同名人数 
    		
    SELECT
    	a.s_name,
    	a.s_sex,
    	count( * ) 
    FROM
    	student a
    	JOIN student b ON a.s_id != b.s_id 
    	AND a.s_name = b.s_name 
    	AND a.s_sex = b.s_sex 
    GROUP BY
    	a.s_name,
    	a.s_sex
    
    
    
    -- 31、查询1990年出生的学生名单
    		
    SELECT
    	s_name 
    FROM
    	student 
    WHERE
    	s_birth LIKE '1990%'
    
    -- 32、查询每门课程的平均成绩，结果按平均成绩降序排列，平均成绩相同时，按课程编号升序排列 
    
    SELECT
    	c_id,
    	ROUND( AVG( s_score ), 2 ) AS avg_score 
    FROM
    	score 
    GROUP BY
    	c_id 
    ORDER BY
    	avg_score DESC,
    	c_id ASC
    
    -- 33、查询平均成绩大于等于85的所有学生的学号、姓名和平均成绩 
    
    SELECT
    	a.s_id,
    	b.s_name,
    	ROUND( avg( a.s_score ), 2 ) AS avg_score 
    FROM
    	score a
    	LEFT JOIN student b ON a.s_id = b.s_id 
    GROUP BY
    	s_id 
    HAVING
    	avg_score >= 85
    	
    -- 34、查询课程名称为"数学"，且分数低于60的学生姓名和分数 
    	
    SELECT
    	a.s_name,
    	b.s_score 
    FROM
    	score b
    	JOIN student a ON a.s_id = b.s_id 
    WHERE
    	b.c_id = ( SELECT c_id FROM course WHERE c_name = '数学' ) 
    	AND b.s_score < 60
    
    -- 35、查询所有学生的课程及分数情况； 
    	
    		
    SELECT
    	a.s_id,
    	a.s_name,
    	SUM( CASE c.c_name WHEN '语文' THEN b.s_score ELSE 0 END ) AS '语文',
    	SUM( CASE c.c_name WHEN '数学' THEN b.s_score ELSE 0 END ) AS '数学',
    	SUM( CASE c.c_name WHEN '英语' THEN b.s_score ELSE 0 END ) AS '英语',
    	SUM( b.s_score ) AS '总分' 
    FROM
    	student a
    	LEFT JOIN score b ON a.s_id = b.s_id
    	LEFT JOIN course c ON b.c_id = c.c_id 
    GROUP BY
    	a.s_id,
    	a.s_name
    
    
     -- 36、查询任何一门课程成绩在70分以上的姓名、课程名称和分数； 
    SELECT
    	a.s_name,
    	b.c_name,
    	c.s_score 
    FROM
    	course b
    	LEFT JOIN score c ON b.c_id = c.c_id
    	LEFT JOIN student a ON a.s_id = c.s_id 
    WHERE
    	c.s_score >= 70
    		
    
    -- 37、查询不及格的课程
    SELECT
    	a.s_id,
    	a.c_id,
    	b.c_name,
    	a.s_score 
    FROM
    	score a
    	LEFT JOIN course b ON a.c_id = b.c_id 
    WHERE
    	a.s_score < 60
    		
    -- 38、查询课程编号为01且课程成绩在80分以上的学生的学号和姓名； 
    SELECT
    	a.s_id,
    	b.s_name 
    FROM
    	score a
    	LEFT JOIN student b ON a.s_id = b.s_id 
    WHERE
    	a.c_id = '01' 
    	AND a.s_score > 80
    -- 39、求每门课程的学生人数 
    SELECT
    	count( * ) 
    FROM
    	score 
    GROUP BY
    	c_id;
    
    -- 40、查询选修"张三"老师所授课程的学生中，成绩最高的学生信息及其成绩
    
    		
    		-- 查询老师id 
    SELECT
    	c_id 
    FROM
    	course c,
    	teacher d 
    WHERE
    	c.t_id = d.t_id 
    	AND d.t_name = '张三'
    		-- 查询最高分（可能有相同分数）
    SELECT
    	MAX( s_score ) 
    FROM
    	score 
    WHERE
    	c_id = '02'
    		-- 查询信息
    SELECT
    	a.*,
    	b.s_score,
    	b.c_id,
    	c.c_name 
    FROM
    	student a
    	LEFT JOIN score b ON a.s_id = b.s_id
    	LEFT JOIN course c ON b.c_id = c.c_id 
    WHERE
    	b.c_id = ( SELECT c_id FROM course c, teacher d WHERE c.t_id = d.t_id AND d.t_name = '张三' ) 
    	AND b.s_score IN ( SELECT MAX( s_score ) FROM score WHERE c_id = '02' )
    
    
    -- 41、查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩 
    SELECT DISTINCT
    	b.s_id,
    	b.c_id,
    	b.s_score 
    FROM
    	score a,
    	score b 
    WHERE
    	a.c_id != b.c_id 
    	AND a.s_score = b.s_score
    	
    
    -- 42、查询每门功成绩最好的前两名 
    		-- 牛逼的写法
    SELECT
    	a.s_id,
    	a.c_id,
    	a.s_score 
    FROM
    	score a 
    WHERE
    	( SELECT COUNT( 1 ) FROM score b WHERE b.c_id = a.c_id AND b.s_score >= a.s_score ) <= 2 
    ORDER BY
    	a.c_id
    
    
    -- 43、统计每门课程的学生选修人数（超过5人的课程才统计）。要求输出课程号和选修人数，查询结果按人数降序排列，若人数相同，按课程号升序排列 
    SELECT c_id, count( * ) AS total FROM score GROUP BY c_id HAVING total > 5 
    ORDER BY
    	total,
    	c_id ASC
    		
    -- 44、检索至少选修两门课程的学生学号 
    		select s_id,count(*) as sel from score GROUP BY s_id HAVING sel>=2
    
    -- 45、查询选修了全部课程的学生信息 
    SELECT
    	* 
    FROM
    	student 
    WHERE
    	s_id IN ( SELECT s_id FROM score GROUP BY s_id HAVING count( * ) = ( SELECT count( * ) FROM course ) )
    
    
    -- 46、查询各学生的年龄
    	-- 按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一
    
    SELECT
    	s_birth,
    	(
    	DATE_FORMAT( NOW( ), '%Y' ) - DATE_FORMAT( s_birth, '%Y' ) - ( CASE WHEN DATE_FORMAT( NOW( ), '%m%d' ) > DATE_FORMAT( s_birth, '%m%d' ) THEN 0 ELSE 1 END ) 
    	) AS age 
    FROM
    	student;
    
    
    -- 47、查询本周过生日的学生
    	select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))=WEEK(s_birth)
    	select * from student where YEARWEEK(s_birth)=YEARWEEK(DATE_FORMAT(NOW(),'%Y%m%d'))
    	
    	select WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))
    
    -- 48、查询下周过生日的学生
    	select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =WEEK(s_birth)
    
    -- 49、查询本月过生日的学生
    
    	select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d')) =MONTH(s_birth)
    	
    -- 50、查询下月过生日的学生
    	select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =MONTH(s_birth)
    
    
    
    -- 51、 查询姓'李'老师的个数
    select COUNT(*) from teacher where teacher.t_name like '李%'